Rational functions with a common iterate - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T19:15:17Z http://mathoverflow.net/feeds/question/115113 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/115113/rational-functions-with-a-common-iterate Rational functions with a common iterate Alexandre Eremenko 2012-12-01T20:59:21Z 2013-04-26T13:39:27Z <p>Let $f$ and $g$ be two rational functions. To avoid trivialities, we suppose that their degrees are at least $2$. We say that they have a common iterate if $f^m=g^n$ for some positive integers $m,n$, where $f^m$ stands for the $m$-th iterate.</p> <blockquote> <p><em>1. Can one describe/classify all such pairs?</em></p> </blockquote> <p>This is probably very hard, and perhaps there exists no simple answer. But here is a simpler question:</p> <blockquote> <p><em>2. Is there an algorithm which finds out whether two rational functions have a common iterate or not ?</em></p> </blockquote> <p>I mean, I give you two rational functions, say with integer coefficients, and you tell me whether they have a common iterate or not. Perhaps using a super-computer...</p> <p>Motivation. J. F. Ritt, (Permutable rational functions. Trans. Amer. Math. Soc. 25 (1923), no. 3, 399-448) gave a complete classification/description of all commuting pairs of rational functions (that is $f(g)=g(f)$)... except when they have a common iterate. I gave a completely different proof of Ritt's theorem, but again it does not apply to the case when $f$ and $g$ have a common iterate (MR1027462).</p> <p>Polynomial pairs (commuting, or with a common iterate) are completely described in MR1501149 Ritt, J. F. On the iteration of rational functions. Trans. Amer. Math. Soc. 21 (1920), no. 3, 348-356, in the very end of this paper.</p> <p>What is the exact relation between permutable pairs and pairs with a common iterate ?</p> <blockquote> <p><em>3. If two functions have a common iterate, must they commute?</em></p> </blockquote> <p>Or perhaps they must, but with explicitly listed exceptions? A positive answer to this will solve problem 2 above. See also my "answer" to <a href="http://mathoverflow.net/questions/48818" rel="nofollow">http://mathoverflow.net/questions/48818</a> for an additional motivation.</p> <p>EDIT. And one more question:</p> <blockquote> <p><em>4. Can one describe commuting functions that have a common iterate?</em></p> </blockquote> <p>This would complete Ritt's description of commuting functions.</p> http://mathoverflow.net/questions/115113/rational-functions-with-a-common-iterate/115147#115147 Answer by Noam D. Elkies for Rational functions with a common iterate Noam D. Elkies 2012-12-02T04:18:15Z 2012-12-04T06:38:31Z <p>Over ${\bf C}$, An easy counterexample to question 3 is $f(x) = x^2$, $g(x) = cx^2$ where $c$ is a nontrivial cube root of unity. Then $f(f(x)) = g(g(x)) = x^4$ but $f$ and $g$ do not commute. There are similar examples for higher iterates.</p> <p><em>[Added later]</em> A more exotic construction yields further examples, some defined over ${\bf Q}$, such as the degree-4 pair $$f(y) = \frac{y^4+18y^2-47}{8y^3}, \phantom{\infty} g(y) = \frac{f-3}{f+1} = \frac{y^4-24y^3+18y^2-27}{y^4+8y^3+18y^2-27}$$ with $f \circ f = g \circ g$ but $f \circ g \neq g \circ f$. This is a "Lattès map" associated to the elliptic curve $E: y^2 = x^3 + 1$: the function $f$ comes from the doubling map $P \mapsto 2P$, and $g$ comes from $P \mapsto 2P+T$ where $T$ is the 3-torsion point $(0,1)$ (as the $(f,g)=(x^2,cx^2)$ example does on the multiplicative group). This elliptic curve yields examples of $f \circ f = g \circ g$ and $f \circ g \neq g \circ f$ with any degree $m^2+mn+n^2$ as long as that's not a multiple of 3, with $f,g \in {\bf Q}(y)$ if $n=0$. Other elliptic curves with complex multiplication yield further examples using the $x$-coordinate rather than the $y$-coordinate, e.g. $f(x) = -x(x^4+6x^2-3)^2 / (3x^4-6x^2-1)^2$ and $g = (f-1)/(f+1)$ from tripling on $y^2=x^3-x$.</p> http://mathoverflow.net/questions/115113/rational-functions-with-a-common-iterate/115198#115198 Answer by Margaret Friedland for Rational functions with a common iterate Margaret Friedland 2012-12-02T20:57:14Z 2012-12-11T18:49:51Z <p>I am replacing my previous incorrect answer by this one. I just learned about a recent preprint by Hexi Ye, </p> <p><a href="http://arxiv.org/pdf/1211.4303.pdf" rel="nofollow">http://arxiv.org/pdf/1211.4303.pdf</a></p> <p>Among other things, he proves, for general $f$ with degree $d \geq 3$, that $\mu_f=\mu_g$ implies that $f$ and $g$ share an iterate (the converse is well known). The symbol $\mu_f$ denotes the unique $f$-invariant measure of maximal entropy for $f$ (and similarly for $g$). He also analyzes generic maps of degree $2$. The proof involves some holomorphic maps from $t \in \mathbb{C}$ to $f_t \in \rm{Rat}_d$, the set of rational functions of degree $d$ (not semigroups, which you point out to be impossible). As far as I can tell at the first glance, he does not seem to address the commutativity question. </p> http://mathoverflow.net/questions/115113/rational-functions-with-a-common-iterate/128785#128785 Answer by Anixx for Rational functions with a common iterate Anixx 2013-04-26T01:41:43Z 2013-04-26T13:39:27Z <p>Regarding the second question the algorithm is as follows:</p> <ol> <li><p>Find the flows (superfunctions) of the both functions in closed form</p></li> <li><p>See if they coincide at integer points.</p></li> </ol> <p>For example, $f(x)=x^2$, $g(x)=x^4$.</p> <p>The flows will be respectively, $C^{2^x}$, $C^{4^x}$.</p> <p>Now we solve</p> <p>$$C^{2^x} = C^{4^y}$$</p> <p>and find $y=x/2$</p> <p>This equation obviously has infinitely many integer solutions.</p> <p>A more complicated case is when $f(x)=\frac{x+1}{x-1}$, $g(x)=\frac{x-1}{x+1}$</p> <p>In this case the flows are:</p> <p>$$f^*(x)=\frac{C \cos \left(\frac{3 \pi x}{4}\right)+\sin \left(\frac{3 \pi x}{4}\right)}{\cos \left(\frac{3 \pi x}{4}\right)-C \sin \left(\frac{3 \pi x}{4}\right)}$$</p> <p>$$g^*(x)=\frac{\left(\left(\sqrt{2}-1\right) C-1\right) (-1)^x+\sqrt{2} C+C+1}{\left(-C+\sqrt{2}+1\right) (-1)^x+C+\sqrt{2}-1}$$</p> <p>Solving equation f*(x) = g*(y) for integer x and y gives $x=4m, y=n$</p>